Question

giải phương trình \(\sqrt{x+3}=\frac{x^3+3x^2+9x}{3x^2+x+3}\)

Answer 1

ĐKXĐ: \(x\ge-3\)

\(\Leftrightarrow x^3+3x^2+9x=\left(3x^2+x+3\right)\sqrt{x+3}\)

\(\Leftrightarrow x^3+3x\left(x+3\right)=3x^2\sqrt{x+3}+\left(x+3\right)\sqrt{x+3}\)

Đặt \(\sqrt{x+3}=a\ge0\) ta được:

\(x^3+3a^2x=3x^2a+a^3\)

\(\Leftrightarrow x^3-3x^2a+3a^2x-a^3=0\)

\(\Leftrightarrow\left(x-a\right)^3=0\)

\(\Leftrightarrow x=a\)

\(\Leftrightarrow x=\sqrt{x+3}\) (\(x\ge0\))

\(\Leftrightarrow x^2-x-3=0\Rightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{13}}{2}\\x=\frac{1-\sqrt{13}}{2}< 0\left(l\right)\end{matrix}\right.\)